# Condition for Total Ordering to be Well-Ordering

## Theorem

Let $A$ be a non-empty class under a total ordering $\preccurlyeq$.

Then $\preccurlyeq$ is a well-ordering on $A$ if and only if the following $3$ conditions are fulfilled:

- $(1): \quad A$ has a smallest element $x_0$

- $(2): \quad$ Every element of $A$ except the greatest (if there is one) has an immediate successor

- $(3): \quad$ For every proper lower section $L$ of $A$, if $L$ has no greatest element, then there is a smallest element of $A$ which is not in $L$.

## Proof

### Sufficient Condition

Let $\preccurlyeq$ be a well-ordering on $A$.

Then by definition of well-ordering, there exists a smallest element of $A$ which we may call $x_0$.

Hence condition $(1)$ is fulfilled.

Let there exist an element $x$ of $A$ such that $x$ has no immediate successor.

Let $S$ be the subclass of $A$ defined as:

- $S = \set {y \in A: x \prec y}$

Suppose $S \ne \O$.

As $S \subset A$ it follows by definition of well-ordering that there exists a smallest element $z$ of $S$.

That is, there exists $z \in A$ such that $x \prec z$ but that there exists no $y \in A$ such that $x \prec y \prec z$.

Hence there can be no element $y \in A$ such that $x \prec y$ and so $S = \O$.

It follows that $x$ must be the greatest element of $A$.

Thus every element of $A$ except the greatest (if there is one) has an immediate successor.

That is, condition $(2)$ is fulfilled.

Let $L$ be a proper lower section of $A$ which has no greatest element.

As $L$ is a proper lower section of $A$, $L$ does not equal $A$.

Therefore the difference $A \setminus L \ne \O$.

But as $A \setminus L \subseteq A$ it follows by the fact that $\preccurlyeq$ is a well-ordering that $A \setminus L$ has a smallest element $z$.

This $z$ is the smallest element of $A$ which is not in $L$.

That is, condition $(3)$ is fulfilled.

$\Box$

### Necessary Condition

Let $\preccurlyeq$ be a total ordering on $A$ which satisfies the following $3$ conditions:

- $(1): \quad A$ has a smallest element $x_0$

- $(2): \quad$ Every element of $A$ except the greatest (if there is one) has an immediate successor

- $(3): \quad$ For every proper lower section $L$ of $A$, if $L$ has no greatest element, then there is a smallest element of $A$ which is not in $L$.

Aiming for a contradiction, suppose $S \subseteq A$ is a subclass of $A$ which does not have a smallest element.

We have from $(1)$ that $x_0 \in A$ is the smallest element of $A$.

Hence if $x_0 \in S$ then $x_0$ is also the smallest element of $S$.

Hence $x_0 \notin S$.

Let $T \subseteq A$ be the class defined as:

- $T := \set {z \in A: \forall y \in S: z \prec y}$

Because $x_0 \in T$ we know that $T$ is non-empty.

Suppose $T$ has a greatest element $g$.

Then, by $(2)$, $g$ has an immediate successor $g'$.

By definition, $g' \in S$.

But by hypothesis $g'$ is not the smallest element of $S$, as $S$ does not have a smallest element.

Then:

- $\exists g'' \in S: g'' \prec g'$

but that means:

- $g \prec g'' \prec g'$ and so $g'$ is not the immediate successor of $g$.

From that contradiction it follows that $S$ has a smallest element.

Otherwise, suppose $T$ does not have a greatest element.

Then by condition $(3)$, $S$ has a smallest element.

Hence both cases contradict the assertion that $S$ has no smallest element.

Hence $S$ does have a smallest element.

As $S$ is arbitrary, it follows that $\preccurlyeq$ is a well-ordering.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Exercise $1.3$